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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Example 47 Find intervals in which the function given by f(๐‘ฅ) =3/10 ๐‘ฅ4 โ€“ 4/5 ๐‘ฅ^3โ€“ 3๐‘ฅ2 + 36/5 ๐‘ฅ + 11 is (a) strictly increasing (b) strictly decreasingf(๐‘ฅ) = 3/10 ๐‘ฅ4 โ€“ 4/5 ๐‘ฅ^3โ€“ 3๐‘ฅ2 + 36/5 ๐‘ฅ + 11 Finding fโ€™(๐’™) fโ€™(๐‘ฅ) = 3/10 ร— 4๐‘ฅ^3 โ€“ 4/5 ร— 3๐‘ฅ^2 โ€“ 3 ร— 2x + 36/5 + 0 fโ€™(๐‘ฅ) = 12/10 ๐‘ฅ^3โ€“ 12/5 ๐‘ฅ^2โ€“ 6x + 36/5 fโ€™(๐‘ฅ) = 6/5 ๐‘ฅ^3โˆ’ 12/5 ๐‘ฅ^2โ€“ 6x + 36/5 fโ€™(๐‘ฅ) = 6(๐‘ฅ^3/5โˆ’(2๐‘ฅ^2)/5โˆ’๐‘ฅ+6/5) fโ€™(๐‘ฅ) = 6((๐‘ฅ^3 โˆ’ 2๐‘ฅ^2โˆ’ 5๐‘ฅ + 6)/5) = 6/5 (๐‘ฅ^3โˆ’2๐‘ฅ^2โˆ’5๐‘ฅ+6) = 6/5 (๐‘ฅโˆ’1)(๐‘ฅ2โˆ’๐‘ฅโˆ’6) = 6/5 (๐‘ฅโˆ’1)(๐‘ฅ2โˆ’3๐‘ฅ+2๐‘ฅโˆ’6) = 6/5 (๐‘ฅโˆ’1)[๐‘ฅ(๐‘ฅโˆ’3)+2(๐‘ฅโˆ’3)] = 6/5 (๐‘ฅโˆ’1)(๐‘ฅ+2)(๐‘ฅโˆ’3) Hence fโ€™(๐‘ฅ) = 6/5 (๐‘ฅโˆ’1)(๐‘ฅ+2)(๐‘ฅโˆ’3) fโ€™(๐‘ฅ) = 6(๐‘ฅ^3/5โˆ’(2๐‘ฅ^2)/5โˆ’๐‘ฅ+6/5) 6/5 (๐‘ฅโˆ’1)(๐‘ฅ+2)(๐‘ฅโˆ’3) = 0 (๐‘ฅโˆ’1)(๐‘ฅ+2)(๐‘ฅโˆ’3) = 0 Hence x = โ€“2 , 1 & 3 Plotting point on real line Thus, we get four disjoint intervals i.e. (โˆ’โˆž,โˆ’2) ,(โˆ’2, 1) ,(1 , 3), (3 , โˆž) โ‡’ f(๐‘ฅ) is strictly decreasing on the interval ๐‘ฅ โˆˆ(โˆ’โˆž,โˆ’๐Ÿ)& (๐Ÿ , ๐Ÿ‘) f(๐‘ฅ) is strictly increasing on the interval ๐‘ฅ โˆˆ(โˆ’๐Ÿ,๐Ÿ) & (๐Ÿ‘ , โˆž)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.